TITLE: What's the expectation of square root of Chi-square variable?
QUESTION [8 upvotes]: Setting
$$X_1,X_2,\ldots \overset{d}{\sim} \mathcal{N}(0,1)$$
$$V = X_1^2 + \ldots + X_n^2$$
Now I would like to find the expectation of $\sqrt{V}$. My biggest problem is I don't know how to write down the density of $\sqrt{V}$. Could someone help me derive it?

REPLY [4 votes]: First note that each $X^{2}_i\sim Gamma(\frac{1}{2},\frac{1}{2})$ for each $i$ (i.e. chi square with df=1) now to find distribution of $\sum_{i=1}^{n}X^2_{i}$ we see that using moment generating functions and fact that each are independent of each other that 
$$M_{\sum_{i=1}^{n}X^2_{i}}=M(t)^{n}=\left(\frac{\frac{1}{2}}{\frac{1}{2}-t}\right)^{\frac{n}{2}}$$
thus $\sum_{i=1}^{n}X^2_{i}\sim Gamma(\frac{n}{2},\frac{1}{2})$ thus $V=\sum_{i=1}^{n}X^2_{i}$ thus 
$$E(\sqrt{V})=\int_0^\infty \frac{\frac{1}{2}^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}v^{\frac{1}{2}}v^{\frac{n}{2}-1}e^{-\frac{1}{2}x}dx=\frac{\frac{1}{2}^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{\Gamma(\frac{n+1}{2})}{\frac{1}{2}^{\frac{n+1}{2}}}\int^\infty_0\frac{\frac{1}{2}^{\frac{n+1}{2}}}{\Gamma(\frac{n+1}{2})}v^{\frac{n+1}{2}-1}e^{-\frac{1}{2}x}dx=\frac{\sqrt{2}\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})}$$